Solve for (x,y) satisfies $$ \left\{\begin{matrix} x-y \in \left [ -1,1 \right ] & \\ 3x-4y \in \left [ 2,3 \right ] \end{matrix}\right. $$
I have tried to solve this by multiplying the first inequality by 4 and subtracting the two inequalities from each other to lead to the solution $ x \in \left [ -7,2 \right ] $ and $ y \in \left [ -6,1 \right ] $. But obviously this is not the correct solution by checking the range of $x-y$
Can someone help me to solve this?
The set of points $(x,y)$ that satisfies the inequalities is a parallelogram:
The simpler way of describing such set is with two inequalities, so there's really nothing to "solve" about the system of inequalities.
this is a system of 4 linear inequality hidden in 2. Its solution is graphical as the value of $x$ and $y$ will depend on each other. You graph $$\begin{cases} x-y\ge -1\\ x-y\le 1\\ 3x-4y\ge 2\\ 3x-4y\le 3 \end{cases}$$
Perhaps, in addition to what has already been proposed, just remember that a line such as the line $d$ with equation $y=x-1$ partitions the plane into three regions: the line $d$ and the two open half-planes defined respectively by $\{(x,y)\in \mathbb R ^2:y-x+1>0\}$ and $\{(x,y)\in \mathbb R ^2: y-x+1<0\}$
